Class 10 Conversion of Solid from One Shape to Another

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♦ Conversion of Solid from One Shape to Another

Conversion of Solid from One Shape to Another

We are sure you would have seen candles. Generally, they are in the shape of a cylinder.
You may have also seen some candles shaped like an animal (see Fig. 13.17).



If you want a candle of any special shape, you will have to heat the wax in a metal container till it becomes completely liquid. Then you will have to pour it into another container which has the special shape that you want.

For example, take a candle in the shape of a solid cylinder, melt it and pour whole of the molten wax into another container shaped like a rabbit. On cooling, you will obtain a candle in the shape of the rabbit.

The volume of the new candle will be the same as the volume of the earlier candle.

This is what we have to remember when we come across objects which are converted from one shape to another, or when a liquid which originally filled one container of a particular shape is poured into another container of a different shape or size, as you see in Fig 13.18.

Q 3250801714

A cone of height 24 cm and radius of base 6 cm is made up of modelling
clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Class 10 Chapter 13 Example 8
Solution:

Volume of cone `= 1/3 xx pi xx 6 xx 6 xx 24 cm^3`

If r is the radius of the sphere, then its volume is ` 4/3 pi r^3`

Since, the volume of clay in the form of the cone and the sphere remains the same, we have

` 4/3 xx pi xx r^3 =1/3 xx pi xx 6 xx 6 xx 24`

i.e., `r^3 = 3 × 3 × 24 = 3^3 × 2^3`

So, `r = 3 × 2 = 6`

Therefore, the radius of the sphere is 6 cm.
Q 3260801715

Selvi’s house has an overhead tank in the shape of a cylinder. This
is filled by pumping water from a sump (an underground tank) which is in the
shape of a cuboid. The sump has dimensions `1.57 m × 1.44 m × 95 cm`. The
overhead tank has its radius `60` cm and height `95` cm. Find the height of the water
left in the sump after the overhead tank has been completely filled with water
from the sump which had been full. Compare the capacity of the tank with that of
the sump. (Use `π = 3.14` )
Class 10 Chapter 13 Example 9
Solution:

The volume of water in the overhead tank equals the volume of the water removed from the sump.

Now, the volume of water in the overhead tank (cylinder) `= πr^2 h`

`= 3.14 × 0.6 × 0.6 × 0.95 m^3`

The volume of water in the sump when full `= l × b × h = 1.57 × 1.44 × 0.95 m^3`

The volume of water left in the sump after filling the tank

`= [(1.57 × 1.44 × 0.95) – (3.14 × 0.6 × 0.6 × 0.95)] m^3 = (1.57 × 0.6 × 0.6 × 0.95 × 2) m^3`


So, the height of the water left in the sump `= ( text (volume of water left in the sump ) )/( l xx b)`

` = (1.57 xx 0.6 xx 0.6 xx 0.95 xx 2)/( 1.57 xx 1.44) m`

`= 0.475 m = 47.5 cm`

Also, ` ( text ( Capacity of tank) ) /( text (Capacity of sump) ) = (3.14 xx 0.6 xx 0.6 xx 0.95)/( 1.57 xx 1.44 xx 0.95 ) = 1/2`

Therefore, the capacity of the tank is half the capacity of the sump.
Q 3280801717

A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of
length 18 m of uniform thickness. Find the thickness of the wire.
Class 10 Chapter 13 Example 10
Solution:

The volume of the rod ` = pi xx (1/2)^2 xx 8 cm^3 = 2 pi cm^3`

The length of the new wire of the same volume `= 18 m = 1800 cm`
If r is the radius (in cm) of cross-section of the wire, its volume` = π × r^2 × 1800 cm^3`

Therefore, `π × r^2 × 1800 = 2π`

i.e., ` r^2 = 1/900`

i.e., ` r = 1/30`

So, the diameter of the cross section, i.e., the thickness of the wire is `1/15` cm ,

i.e., `0.67mm` (approx.).
Q 3200801718

A hemispherical tank full of water is emptied by a pipe at the rate of `3 4/7`
litres per second. How much time will it take to empty half the tank, if it is `3m` in
diameter? (Take ` pi = 22/7 ` )
Class 10 Chapter 13 Example 11
Solution:

Radius of the hemispherical tank` = 3/2 m`

Volume of the tank `= 2/3 xx 22/7 xx (3/2)^3 m^3 = 99/14 m^3`

So, the volume of the water to be emptied `= 1/2 xx 99/14 m^3 = 99/28 xx 1000 ` liters

` = 99000/28` liters

Since , `25/7` litres of water is emptied in 1 second, `99000/28` litres of water will be emptied
in `99000/28 xx 7/25` seconds, i.e., in 16.5 minutes.

 
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